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Let $(X,\mu,\mathcal{F})$ be a probability space. The paper Equiconvergence of Martingales by Edward Boylan introduced a pseudometric on sub-$\sigma$-fields (sub-$\sigma$-algebras) of $\mathcal{F}$ as follows:

$\rho(\mathcal{G},\mathcal{H}) := \sup_{A\in \mathcal{G}} \inf_{B\in \mathcal{H}} \mu(A \triangle B) + \sup_{B\in \mathcal{H}} \inf_{A\in \mathcal{G}} \mu(A \triangle B)$

where $A \triangle B$ is symmetric difference.

It seems to be called the Hausdorff pseudometric on $\sigma$-fields in later papers. (Does anyone know why?) Further, if we only consider a $\mu$-complete $\sigma$-fields then $\rho$ is a metric. Also, the paper shows $\rho$ is complete.

Is this metric $\rho$ separable---assuming, say, $X=[0,1]$ and $\mu$ is the Lebesgue measure?

My guess is that it is not, but I cannot off-hand come up with a witnessing set to show this. Considering the paper is 40 years old, I imagine this might be well-known. And if it is not separable, then my follow up question is this?

Is there a known separable, complete metric on the space of $\mu$-complete sub-$\sigma$-fields?

For reference, I found the following list online, compiled by Dave L. Renfro, of papers dealing with metrics on $\sigma$-fields (listed in Chronological order). I quickly looked though these papers and didn't find what I was looking for, but maybe I missed something.

  1. Edward S. Boylan, "Equiconvergence of martingales",
    Annals of Mathematical Statistics 42 (1971), 552-559. [MR 44 #7603; Zbl 218.60049]

  2. Jacques Neveu, "Note on the tightness of the metric on the set of complete sub sigma-algebras of a probability space", Annals of Mathematical Statistics 43 (1972), 1369-1371. [MR 48 #5133; Zbl 241.60036]

  3. Hirokichi Kudo, "A note on the strong convergence of sigma-algebras", Annals of Probability 2 (1974), 76-83. [MR 51 #6900; Zbl 275.60007]

  4. Lothar Rogge, "Uniform inequalities for conditional expectations", Annals of Probability 2 (1974), 486-489. [MR 50 #14858; Zbl 285.28010]

  5. Louis H. Blake, "Some further results concerning equiconvergence of martingales", Revue Roumaine de Mathématiques Pures et Appliquées 28 (1983), 927-932. [MR 86i:60130; Zbl 524.60029]

  6. Hari G. Mukerjee, "Almost sure equiconvergence of conditional expectations", Annals of Probability 12 (1984), 733-741. [MR 86c:28012; Zbl 557.28001]

  7. Beth Allen, "Convergence of sigma-fields and applications to mathematical economics", pp. 161-174 in Gerald Hammer and Diethard Pallaschke (editors), SELECTED TOPICS IN OPERATIONS RESEARCH AND MATHEMATICAL ECONOMICS (Proceedings, Karlsruhe, West Germany, 22-25 August 1983), Lecture Notes in Economics and Mathematical Systems #226, Springer-Verlag, 1984. [MR 86f:90029; Zbl 547.28001]

  8. Dieter Landers and Lothar Rogge, "An inequality for the Hausdorff-metric of sigma-fields", Annals of Probability 14 (1986), 724-730. [MR 87h:60006; Zbl 597.60003]

  9. Abdallah M. Al-Rashed, "On countable unions of sigma algebras", Journal of Karachi Mathematical Association 8 (1986), 57-63. [MR 88f:28001; Zbl 639.28001]

  10. Maxwell B. Stinchcombe, "A further note on Bayesian information topologies", Journal of Mathematical Economics 22 (1993), 189-193. [MR 93k:60011; Zbl 773.90016]

  11. Timothy Van Zandt, "The Hausdorff metric of sigma-fields and the value of information", Annals of Probability 21 (1993), 161-167. [MR 94d:62012; Zbl 777.62007]

  12. Xikui Wang, "Completeness of the set of sub-sigma-algebras", International Journal of Mathematics and Mathematical Sciences 16 (1993), 511-514. [MR 94f:28002; Zbl 782.28001]

  13. Zvi Artstein, "Compact convergence of sigma-fields and relaxed conditional expectation", Probability Theory and Related Fields [= Zeitschrift für Wahrscheinlichkeits- theorie] 120 (2001), 369-394. [MR 2002g:28003; Zbl 992.28001]

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It is called the Hausdorff pseudometric because it is an instance of Haudsorff's construction starting with pseudometric $\mu(A \triangle B)$. In general, Hausdorff's construction starts with a pseudometric and constructs a new pseucometric on subsets of the original space. See here en.wikipedia.org/wiki/Hausdorff_distance . –  Gerald Edgar Nov 3 '11 at 0:25
    
As a first step, what is the cardinality of the set of complete $\sigma$-algebras? If it isn't $2^{\aleph_0}$, then that's certainly an obstruction. –  Nate Eldredge Nov 3 '11 at 4:52
    
@Gerald Edgar, thanks I thought it might be something like that. –  Jason Rute Nov 3 '11 at 14:12
    
@Nate Eldredge, I think it is size continuum as follows: If $\rho(\mathcal{G},\mathcal{H})\neq 0$, then $f \mapsto E[f \mid \mathcal{G}]$ and $f \mapsto E[f \mid \mathcal{H}]$ are different operators. But in $L^2$ these operators are continuous linear transformations of which there are only continuum many (correct?). –  Jason Rute Nov 3 '11 at 14:18

2 Answers 2

up vote 3 down vote accepted

Take a sequence $A_n$ of independent sets of measure $1/2$. Given two different subsets $B$ and $C$ of natural numbers, suppose WLOG that there is an $n$ in $B\sim C$. Now $\mu(A_n\Delta A) = 1/2$ for all sets $A$ which are independent of $A_n$, so the distance from the sigma algebra generated by $(A_n)_{n\in B}$ to the sigma algebra generated by $(A_n)_{n\in C}$ is at least $1/2$. This shows that the density character of your space is at least the continuum.

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Thanks! I assume $B\sim C$ is set subtraction. I feel a bit silly for not noticing this construction. Also, by my comment to Nate above, the space has size continuum, so the density character is exactly the continuum. –  Jason Rute Nov 3 '11 at 14:30
    
That's right, Jason. –  Bill Johnson Nov 3 '11 at 18:00

I suspect that the following is a dense set:

For each $n\in\mathbb{N}$ take all sub-algebras of the finite sigma-algebra generated by intervals of the form $[i/2^n,(i+1)/2^n)$, $i=0,\ldots,2^n-1$.

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Unfortunately not. (I had thought this at first too.) By Bill Johnson's answer, every pair in your sequence has distance 1/2. (So this sequence doesn't even converge in this metric to the Borel $\sigma$-algebra which it generates.) A similar example is in the original paper. (I just didn't see at the time how to extend it to a collection of size continuum as Bill did.) –  Jason Rute Nov 3 '11 at 14:39
    
I see. I was sure wrong. –  Yuri Bakhtin Nov 3 '11 at 18:28

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